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Let $(X,d)$ be a locally compact metric space. Then for each $x \in X$ $\exists$ $\epsilon_x > 0$ such that $B[x;\epsilon_x] = \{y \in X : d(x,y) \leq \epsilon_x \}$ is compact.

How do I proceed to prove it? Please help me in this regard.

Thank you very much.

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Pretty simple!

Let us choose $x \in X$ arbitrarily. Since $X$ is locally compact so $\exists$ a compact neighbourhood $C_x$ of $x$ in $X$ i.e. $\exists$ $\epsilon >0$ such that $x \in B(x;\epsilon) \subset C_x$. Consider $0 < \epsilon_x < \epsilon$ then clearly $B[x;\epsilon_x] \subset B(x;\epsilon) \subset C_x$. Now $B[x;\epsilon]$ (being a closed subset of a compact set $C_x$) is compact. Which proves your claim.

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  • $\begingroup$ Actually it is a necessary and sufficient condition for locally compactness. $\endgroup$ – Dbchatto67 Aug 19 '18 at 6:56

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